The path of a shot put released at an angle of 35° can be modeled by y=−0.01x2+0.7x+6, where x is the horizontal distance (in feet) and y is the vertical distance (in feet). Find and interpret the coordinates of the vertex.

The coordinates of the vertex are ( , )

When the shot put is at its highest point, it is ( ) feet from its starting point and ( ) feet off the ground.

The distance from the highest point of the shoot put to its initial position is the distance of the line joining the vertex of the parabola and the initial position of shoot put.

Here you find the distance between point A(0,6) and B(35,18.3)

Apply the distance formula;

[tex]d=\sqrt{(x_2-x_1)^{2}+(y_2-y_1)^{2} } \\\\\\d=\sqrt{35-0)^2+(18.3-6)^2} \\\\\\d=\sqrt{35^2+12.3^2} \\\\\\d=\sqrt{1376.29} \\\\\\d=37.098\\\\d=37.1[/tex]

MrRoyal MrRoyal · Answer 2 · 2021-10-02T21:53:56+02:00

The vertex of an equation is the maximum or minimum point of the graph

The coordinates of the vertex is: (70,6)
When the shot put is at its highest point, it is 37.1 feet from its starting point and 35 feet off the ground.

The equation is given as:

[tex]\mathbf{y = -0.01x^2 + 0.7x + 6}[/tex]

For a quadratic equation:

[tex]\mathbf{y = ax^2 + bx + c}[/tex]

The coordinates of the parabola is:

[tex]\mathbf{(h,k)}[/tex]

Where:

[tex]\mathbf{h = -\frac{b}{2a}}[/tex]

[tex]\mathbf{k = f(h)}[/tex]

In [tex]\mathbf{y = -0.01x^2 + 0.7x + 6}[/tex]

[tex]\mathbf{b = 0.7}[/tex]

[tex]\mathbf{a = -0.01}[/tex]

So:

[tex]\mathbf{h = -\frac{b}{2a}}[/tex]

[tex]\mathbf{h = \frac{-0.7}{-2 \times 0.01}}[/tex]

[tex]\mathbf{h = 35}[/tex]

Recall that:

[tex]\mathbf{k = f(h)}[/tex]

So, we have:

[tex]\mathbf{k = f(35)}[/tex]

This gives:

[tex]\mathbf{k = -0.01\times 35^2 + 0.7 \times 35 + 6}[/tex]

[tex]\mathbf{k = 18.25}[/tex]

Hence, the coordinates of the vertex is: (35,18.25)

The starting point is when [tex]\mathbf{x = 0}[/tex]

So, we have:

[tex]\mathbf{y = -0.01\times 0^2 + 0.7 \times 0 + 6}[/tex]

[tex]\mathbf{y = 6}[/tex]

The coordinates of the starting point is (0,6)

The distance between the starting point and the vertex is:

[tex]\mathbf{d =\sqrt{(x_2 -x_1)^2 + (y_2 - y_1)^2}}[/tex]

[tex]\mathbf{d =\sqrt{(35 -0)^2 + (18.25 - 6)^2}}[/tex]

[tex]\mathbf{d =\sqrt{1375.0625}}[/tex]

[tex]\mathbf{d =37.1}[/tex]

So, the interpretation is that:

When the shot put is at its highest point, it is 37.1 feet from its starting point and 35 feet off the ground.